\newproblem{lay:4_6_1}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 4.6.1}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Assume
	\begin{center}
		$A=\begin{pmatrix}1 & -4 & 9 & -7 \\ -1 & 2 & -4 & 1 \\ 5 & -6 & 10 & 7\end{pmatrix} \sim
		   \begin{pmatrix} 1 & 0 & -1 & 5 \\ 0 & -2 & 5 & -6 \\ 0 & 0 & 0 & 0\end{pmatrix}=B$
	\end{center}
	Without calculations list $\mathrm{Rank}\{A\}$ and $\dim\{\mathrm{Nul}\{A\}\}$. Then find bases for $\mathrm{Col}\{A\}$, $\mathrm{Row}\{A\}$, and $\mathrm{Nul}\{A\}$.
}{
  % Solution
	The rank of $A$ is the dimension of the column space of $A$ that is given by the number of pivot columns of $A$. This is the same as the number of pivot columns of $B$, that is, 2
	(first and second columns). The dimension of the null space of $A$, thanks to the rank theorem, can be calculated as the number of columns of $A$ minus its rank,
	in this case, $4-2=2$.
	
	The basis of $\mathrm{Col}\{A\}$ is given by the pivot columns of $A$, that are the same as the pivot columns of $B$:
	\begin{center}
		$\mathrm{Basis}\{\mathrm{Col}\{A\}\}=\left\{\begin{pmatrix}1\\-1\\5\end{pmatrix},\begin{pmatrix}-4\\2\\-6\end{pmatrix}\right\}$
	\end{center}
	
	The space spanned by the rows of $B$ is the same as the space spanned by the rows of $A$. So a basis for the row space of $A$ is
	\begin{center}
		$\mathrm{Basis}\{\mathrm{Row}\{A\}\}=\left\{\begin{pmatrix}1 \\ 0 \\ -1 \\ 5\end{pmatrix},\begin{pmatrix}0 \\ -2 \\ 5 \\ -6\end{pmatrix}\right\}$
	\end{center}
	
	For the null space of $A$ we write the equations implied by the two rows of $B$
	\begin{center}
		$\begin{array}{c}x_1=x_3-x_4 \\ -2x_2=-5x_3+6x_4\end{array} \Rightarrow \begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}=x_3\begin{pmatrix}1\\ \frac{5}{2}\\1\\0\end{pmatrix}+
		   x_4\begin{pmatrix}-1\\-3\\0\\1\end{pmatrix}$
	\end{center}
	So the basis of $\mathrm{Nul}\{A\}\}$ is given by
	\begin{center}
		$\mathrm{Basis}\{\mathrm{Row}\{A\}\}=\left\{\begin{pmatrix}1\\ \frac{5}{2}\\1\\0\end{pmatrix},\begin{pmatrix}-1\\-3\\0\\1\end{pmatrix}\right\}$
	\end{center}
}
\useproblem{lay:4_6_1}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
